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Chapter 1: Problem 45

How many minutes does it take light from the sun to reach Earth? (The distance from the sun to Earth is 93 million \(\mathrm{mi} ;\) the speed of light \(=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\).)

Short Answer

Expert verified
It takes approximately 8315 minutes for light from the Sun to reach the Earth.

Step by step solution

01

Convert distance from miles to meters

1 mile is approximately equal to 1609.34 meters. Therefore, 93 million miles would be equal to \(93,000,000 \times 1609.34 = 1.4966862 \times 10^{14}\) meters.
02

Use speed-distance-time formula to solve for time

The formula for speed is speed = distance/time. This can be rearranged to solve for time: time = distance/speed. Substituting the given values, calculate the time it takes for light to travel from the Sun to the Earth: \( time = \frac{1.4966862 \times 10^{14}}{3.00 \times 10^{8}} = 4.988954 \times 10^{5} \) seconds.
03

Convert time from seconds to minutes

There are 60 seconds in a minute. Therefore, to convert seconds to minutes, divide the result from Step 2 by 60: \( \frac{{4.988954 \times 10^{5}}}{{60}} = 8314.92 \) minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance Conversion
Understanding how to convert units of distance is a vital skill in physics and everyday situations alike. In the context of light travel from the Sun to the Earth, converting the distance from miles to meters is necessary because the speed of light is commonly expressed in meters per second (\( m/s \)).

To perform this conversion, we use the fact that 1 mile is approximately equal to 1609.34 meters. Consequently, if we have a distance in miles, we multiply by this conversion factor to find the equivalent distance in meters. For example, if the Sun is 93 million miles away, the calculation would look like this: \( 93,000,000 \text{ miles} \times 1609.34 \text{ m/mile} \). The result is a much larger number in meters, because meters are a smaller unit than miles, which reflects the incredible vastness of space that light can travel.
Speed-Distance-Time Formula
The relationship between speed, distance, and time is a foundational concept in physics, commonly known as the speed-distance-time formula. This relationship is given by the equation \( \text{speed} = \frac{\text{distance}}{\text{time}} \).

However, when we need to find the time it takes for something to happen, like how long it takes light to travel from the Sun to the Earth, we rearrange the formula to \( \text{time} = \frac{\text{distance}}{\text{speed}} \). By plugging in the distance in meters and the speed of light in meters per second, we can calculate the time it takes for light to cover that distance. This equation helps us understand that the higher the speed or the shorter the distance, the less time it will take for an object to travel. Conversely, the greater the distance or the slower the speed, the more time it will take.
Light Travel Time Calculation
The travel time calculation for light is a fascinating application of the speed-distance-time formula. Light moves at an incredibly fast, constant speed of approximately \(3.00 \times 10^8 m/s\) in a vacuum, like space. To determine how long it takes for light to travel a given distance, we divide the distance the light has to travel by its speed.

After calculating the travel time in seconds, we often convert it into more understandable units, such as minutes or hours. For instance, by dividing the time in seconds by 60, we convert it to minutes. This allows us to appreciate the astronomical distances in our universe on a more human scale, such as understanding that light from the Sun takes just over eight minutes to reach Earth 鈥� an astounding fact that illustrates both the vastness of space and the incredible speed at which light travels.

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Most popular questions from this chapter

Percent error is often expressed as the absolute value of the difference between the true value and the experimental value, divided by the true value: percent error \(=\frac{\mid \text { true value }-\text { experimental value } \mid}{\mid \text { true value } \mid} \times 100 \%\) The vertical lines indicate absolute value. Calculate the percent error for the following measurements: (a) The density of alcohol (ethanol) is found to be \(0.802 \mathrm{~g} / \mathrm{mL}\). (True value: \(0.798 \mathrm{~g} / \mathrm{mL}\).) (b) The mass of gold in an earring is analyzed to be \(0.837 \mathrm{~g}\). (True value: \(0.864 \mathrm{~g}\).)

A \(250-\mathrm{mL}\) glass bottle was filled with \(242 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) and tightly capped. It was then left outdoors overnight, where the average temperature was \(-5^{\circ} \mathrm{C}\). Predict what would happen. The density of water at \(20^{\circ} \mathrm{C}\) is \(0.998 \mathrm{~g} / \mathrm{cm}^{3}\) and that of ice at \(-5^{\circ} \mathrm{C}\) is $0.916 \mathrm{~g} / \mathrm{cm}^{3}.

The following procedure was used to determine the volume of a flask. The flask was weighed dry and then filled with water. If the masses of the empty flask and filled flask were \(56.12 \mathrm{~g}\) and \(87.39 \mathrm{~g},\) respectively, and the density of water is \(0.9976 \mathrm{~g} / \mathrm{cm}^{3}\), calculate the volume of the flask in \(\mathrm{cm}^{3}\).

Express the answers to the following calculations in scientific notation: (a) \(0.0095+\left(8.5 \times 10^{-3}\right)\) (b) \(653 \div\left(5.75 \times 10^{-8}\right)\) (c) \(850,000-\left(9.0 \times 10^{5}\right)\) (d) \(\left(3.6 \times 10^{-4}\right) \times\left(3.6 \times 10^{6}\right)\)

In determining the density of a rectangular metal bar, a student made the following measurements: length, \(8.53 \mathrm{~cm} ;\) width, \(2.4 \mathrm{~cm} ;\) height, \(1.0 \mathrm{~cm}\) mass, 52.7064 g. Calculate the density of the metal to the correct number of significant figures.
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